CBSE Class 12 Math 2013 Solved Paper

© examsnet.com
Question : 12
Total: 29
Find the value of the following:
tan
1
2
|sin−1
2x
1+x2
+cos−1
1−y2
1+y2
|
, |x| < 1, y > 0 and xy < 1
OR
Prove that tan−1(
1
2
)
+ tan−1(
1
5
)
+ tan−1(
1
8
)
=
Ï€
4
Solution:  
We know that:
sin−1
2x
1+x2
= 2 tan−1 x for |x| ≤ 1 .. (1)
cos−1
1−y2
1+y2
= 2 tan−1 y for y = 0 ... (2)
∴ sin−1
2x
1+x2
+ cos−1
1−y2
1+y2
= 2tan−1 x + 2 tan−1 y
⇒ tan
1
2
|sin−1
2x
1+x2
+cos−1
1−y2
1+y2
|

= tan
1
2
(2 tan−1 x + 2 tan−1 y)
= tan (tan−1x+tan−1y)
= tan (tan−1
x+y
1−xy
)

[Since tan−1x+tan−1y = tan−1
x+y
1−xy
, for xy < 1]
=
x+y
1−xy

OR
We know that:
tan−1x+tan−1y = tan−1
x+y
1−xy
, for xy < 1
We have:
tan−1(
1
2
)
+ tan−1(
1
5
)
+ tan−1(
1
8
)

= |tan−1(
1
2
)
+ tan−1(
1
5
)
|
+ tan−1(
1
8
)

= tan−1(
1
2
+
1
5
1−
1
2
×
1
5
)
+ tan−1(
1
8
)
(Since
1
2
×
1
5
<1)
= tan−1(
7
9
)
+tan−1(
1
8
)

= tan−1
7
9
+
1
8
1−
7
9
×
1
8

= tan−1
56+9
72−7
(Since
7
9
×
1
8
< 1)
= tan−1
65
65
= tan−1 1 =
Ï€
4

Hence, tan−1(
1
2
)
+ tan−1(
1
5
)
+ tan−1(
1
8
)
=
Ï€
4
© examsnet.com
Go to Question: